GENERALIZING THE TITCHMARSH DIVISOR PROBLEM
ADAM TYLER FELIX
Abstract. Let a be a natural number different from 0. In 1963, Linnik proved the following
unconditional result about the Titchmarsh divisor problem
X
x log log x
d(p − a) = cx + O
log x
p≤x
where c is a constant dependent on a. Titchmarsh proved the above result assuming GRH for
Dirichlet L-functions in 1931.
We establish the following asymptotic relation:
X
p−a
x
d
= Ck x + O
k
log x
p≡a
p≤x
mod k
where Ck is a constant dependent on k and a and the implied constant is dependent on k.
We also apply it a question related to Artin’s conjecture for primitive roots.
1. Introduction
Let d(n) denote the number of positive divisors of n ∈ N. Let p denote a prime number.
Then, the Titchmarsh divisor problem [2, §9.3] is concerned with the following summation:
X
d(p − a)
p≤x
where a is a fixed non-zero integer. The above summation was first investigated by Titchmarsh
[15], who proved the following theorem:
Theorem 1.1 (Titchmarsh). Suppose the Generalized Riemann Hypothesis holds for Dirichlet
L-functions. Then
X
ζ(2)ζ(3) Y
p
x log log x
(1.1)
d(p − a) =
1− 2
x+O
ζ(6)
p −p+1
log x
a<p≤x
p|a
as x → ∞.
In 1961, Linnik [10] proved the above asymptotic formula with his dispersion method,
thereby eliminating the use of the generalized Riemann hypothesis from the above theorem.
1991 Mathematics Subject Classification. 11N37, 11N36, 11N13.
Key words and phrases. Titchmarsh divisor problem, primes in arithmetic progression, Artin’s conjecture.
Research of the first author supported by an NSERC PGS-D scholarship.
1
2
ADAM TYLER FELIX
Rodriquez [14] and Halberstam [7] independently showed that the above can be proven unconditionally using the Bombieri-Vinogradov Theorem. In fact, Fouvry [6, Corollaire 2], and
Bombieri, Friedlander and Iwaniec [1, Corollary 1] have shown for any A > 1,
X
x
(1.2)
d(p − a) = cx + c1 li(x) + O
(log x)A
p≤x
where c and c1 are effectively computable constants dependent on a and the implied constant
depends only on a and A, and li(x) is the usual logarithmic integral.
1.1. Notation. The letter p will denote a prime number. The letter k will denote a positive
integer. The function d(n) will denote the number of positive divisors of n ∈ N. The Euler
totient function, which will be denoted by ϕ(n) with n ∈ N, is the number of coprime residue
classes in Z/nZ. Let a, k ∈ N with gcd(a, k) = 1 and x ∈ R with x ≥ 1. Then, by π(x; k, a)
denote by the number #{p ≤ x : p ≡ a mod k}, and by ψ(x; k, a) we denote by the summation
X
(1.3)
ψ(x; k, a) :=
Λ(n)
n≤x
n≡a mod k
where Λ(n) = log p if n = pα for some prime p and α ∈ N and Λ(n) = 0 otherwise. That is,
Λ is the von Mangoldt function. For p - a, define fa (p) = min{d ∈ N : ad ≡ 1 mod p} and for
p|a, define fa (p) = ∞. For p - a, define ia (p) = fp−1
and for p|a, define ia (p) = 0. We call
a (p)
fa (p) the order of a modulo p and ia (p) the index of a modulo p. The logarithmic integral,
denoted by li(x), is the integral
Z x
dt
(1.4)
li(x) :=
.
2 log t
Let f : R → C and g : R → R≥0 . We will write f (x) = O(g(x)) if there is a constant C
such that |f (x)| ≤ Cg(x) for all x ∈ R. Equivalently, we will write f (x) g(x) for the same
relation. We will write f (x) g(x) if f (x) g(x) f (x). Here f has codomain R≥0 . We
will write f (x) ∼ g(x) if
(1.5)
f (x)
= 1.
x→∞ g(x)
lim
We will write f (x) = o(g(x)) if
(1.6)
f (x)
= 0.
x→∞ g(x)
lim
1.2. Statement of Results. We wish to consider
X
p−1
(1.7)
d
k
p≤x
p≡1 mod k
where k ∈ N is fixed. We will also consider the above summation with d((p − 1)/k) replaced
by d((p − a)/k) as p ranges over p ≤ x with p ≡ a mod k. However, the case a = 1 has
GENERALIZING THE TITCHMARSH DIVISOR PROBLEM
3
applications to Artin’s conjecture for primitive roots.
In §2, we will prove the following theorems:
Theorem 1.2. For any k ∈ N, k > 1 be an integer, and let a ∈ Z such that gcd(a, k) = 1,
and A > 0. We have the following results uniformly in k ≤ (log x)A+1 :
X
ck
log k
1
= log x + O
(1.8)
ϕ(kd)
k
k
d≤x
gcd(a,d)=1
where the O-constant is dependent only on a, and
X
ck
x(log k)(1 + ck )
x
p−a
= x+O
+O
(1.9)
d
k
k
k
log
x
(log x)A
p≤x
p≡a mod k
where
ck := ck (a) =
X
w≥1
gcd(w,a)=1
Y
=
1+
p-a
p-k
(1.10)
µ2 (w) gcd(w, k)
wϕ(w)
1
p(p − 1)
Y
1+
p|k
1
p−1
Y
1
1−
p
p|a
Y
ζ(2)ζ(3) Y
p
p−1
=
1− 2
1+ 2
ζ(6)
p −p+1
p −p+1
p|a
p|k
and the first O-constant is absolute and the second O-constant is dependent only on a and A.
In §3, we will prove
Lemma 1.1. Let a = 1. Let ck := ck (1) be defined as above. Then, we have
X µ(k)ck
1
=1+O
(1.11)
.
2
k
x
k≤x
and
Theorem 1.3. We have
(1.12)
X ϕ(k)
k≤x
k
π(x; k, 1) = x + O
x
log x
as x → ∞.
These above theorems will then give an application to a problem related to Artin’s conjecture
for primitive roots:
4
ADAM TYLER FELIX
Theorem 1.4. Let y be a function of x such that logx x = o(y). Then,
x
1 XX 1
= log x + O(log log x) + O
.
(1.13)
y a≤y p≤x fa (p)
y
and
Theorem 1.5. Let y be a function of x such that logx x = o(y). Then,
1 XX
1
ζ(2)ζ(3)
x
(1.14)
=
log x + c1 log log x + O
y a≤y p≤x ϕ(fa (p))
ζ(6)
y
where
!
log p
−2
+ 2γ − 1
2−p+1
p
p prime
ζ(2)ζ(3)
c1 =
ζ(6)
(1.15)
X
where γ is Euler’s constant:
(1.16)
!
X 1
− log x .
n
1≤n≤x
γ := lim
x→∞
2. Proof of Theorem 1.2
Let
√
if n ∈ Z
.
otherwise
(
1
δ(n) :=
0
(2.1)
So, we have
(2.2)
X
d(n) = 2
1 + δ(n).
d|n
√
d< n
Then,
X
d
p≤x
p≡a mod k
p−a
k
=2
X
X
√ p−a
p≤x
d≤
k
p≡a mod k
d| p−a
k
X
1+
δ
p≤x
p≡a mod k
p−a
k
!
(2.3)
=2
X
√x
d≤2
π(x; kd, a) + O
X
δ(n) .
n≤x
k
Now,
X
δ(n) = #{n ≤ x : n = m2 for some m ∈ N}
n≤x
(2.4)
= #{m2 ≤ x} ≤
√
x.
GENERALIZING THE TITCHMARSH DIVISOR PROBLEM
5
So, we have
X
(2.5)
d
p≤x
p≡a mod k
p−a
k
=2
X
√x
d≤2
π(x; kd, a) + O
√ x .
k
However, since π(x; kd, a) ≤ 1 if gcd(a, d) > 1, we have

X
√x
d≤2
π(x; kd, a) =
k
(2.6)
X
√
d≤2 x
k
gcd(d,a)=1
li(x)
li(x)
π(x; kd, a) −
+
ϕ(kd) ϕ(kd)


+O


X
√
d≤2 x
k
gcd(d,a)6=1


1



 X
X
1
li(x) 


+ O 
π(x; kd, a) −
= li(x)

ϕ(kd) 

√ ϕ(kd)
d≤2√ xk
d≤2 x
k
gcd(d,a)=1
gcd(d,a)=1
√
+ O( x).
Hence, we have


 X

X
X
1
li(x)


π(x; kd, a) = li(x)
+ O 
π(x; kd, a) −


ϕ(kd)
ϕ(kd)

√
√x
√x
d≤2 xk
d≤2 k
d≤2 k
gcd(d,a)=1
gcd(d,a)=1
√
+ O( x)
X
1
x
(2.7)
= li(x)
+O
,
(log x)A
√ x ϕ(kd)
d≤2 k
gcd(d,a)=1
where we have used [1, Theorem 9] which states the following: choose ε > 0. If we assume
1
k < x 10 −ε (which is true since, in our case, k < (log x)A+1 ), then, there exists C := C(A) such
that for any Q ≤ x/k(log x)C we have
X X
x
x
(2.8)
ψ(x; nm, a) −
ϕ(qr) (log x)A
n≤Q
m≤k
gcd(a,m)=1 gcd(n,a)=1
where the implied constant depends
on√at most ε, a, and A, and
only on
p
√ C is dependent
x
(A+1)/2
C
A. We note that k(log x)B > x/k as k < (log x)
< x/(log x) . Therefore, by
partial summation and concerning ourselves only with m = k, which satisfies gcd(a, k) = 1 by
6
ADAM TYLER FELIX
hypothesis, in the above summation, we have
X
(2.9)
√x
d≤2
X
π(x; kd, a) = li(x)
√x
1
+O
ϕ(kd)
x
(log x)A
.
d≤2 k
gcd(d,a)=1
k
So, we just need to deal with
X
d≤x
gcd(d,a)=1
1
.
ϕ(kd)
We will evaluate this summation by using [2, Exercises 6.3 and 6.4] as a guide. We have
X
d≤x
gcd(d,a)=1
kd
=
ϕ(kd)
=
(2.10)
=
d≤x
w|kd
gcd(d,a)=1
X µ2 (w)
ϕ(w)
w≤kx
X
w≤kx
gcd(w,a)=1
X µ2 (w)
=
ϕ(w)
ϕ(w)
w≤kx
X µ2 (w)
X
X
d≤x
µ2 (w)
ϕ(w)
1
d≤x
w|kd
gcd(d,a)=1
X
1=
w
|d
gcd(w,k)
gcd(d,a)=1
X
w≤kx
µ2 (w)
ϕ(w)
1
d≤x
w
|d
gcd(w,k)
gcd(d,a)=1
w
,a)=1
gcd( gcd(w,k)
X
X
1
x gcd(w,k)
d≤
w
gcd(d,a)=1
w
,a
gcd(w,k)
w
,a
gcd(w,k)
since gcd(a, k) = 1 implies gcd
= gcd(w, a), and gcd
= 1 implies
w
gcd gcd(w,k)
d, a = gcd(d, a). Also, since (mutatis mutandis the proof of [11, Exercise 1.5.8])
(2.11)
X
d≤y
gcd(d,a)=1
1=
ϕ(a)
y + O(d(a)),
a
GENERALIZING THE TITCHMARSH DIVISOR PROBLEM
7
we have
X
d≤x
gcd(d,a)=1
kd
=
ϕ(kd)
X
w≤kx
gcd(w,a)=1
ϕ(a)
x
=
a
ϕ(a)
=x
a
(2.12)
µ2 (w)
ϕ(w)
X
w≤kx
gcd(w,a)=1
X
w≥1
gcd(w,a)=1
ϕ(a) x gcd(w, k)
+ O(d(a))
a
w
X 1
µ2 (w) gcd(w, k)
+ O d(a)
wϕ(w)
ϕ(w)
w≤kx
!
X µ2 (w) gcd(w, k)
µ2 (w) gcd(w, k)
+O x
wϕ(w)
wϕ(w)
w>kx
!
+ O (d(a) log(kx)) .
However,
X µ2 (w) gcd(w, k)
X µ2 (w)
1
k log(kx)
1
≤k
+O
.
wϕ(w)
wϕ(w)
x
(kx)2
x
w>kx
w>kx
(2.13)
So
X
d≤x
gcd(d,a)=1
(2.14)
kd
ϕ(a)
=x
ϕ(kd)
a
=x
X
w≥1
gcd(w,a)=1
X µ2 (w) gcd(w, k)
µ2 (w) gcd(w, k)
+O x
wϕ(w)
wϕ(w)
w>kx
+ O (log(kx))
ϕ(a) X µ2 (w) gcd(w, k)
a
w≥1
gcd(w,a)=1
wϕ(w)
+ O(log(kx)).
We notice that
(2.15)
X
w≥1
gcd(w,a)=1
X µ2 (w)
µ2 (w) gcd(w, k)
≤k
,
wϕ(w)
wϕ(w)
w≥1
which converges absolutely for k fixed. Define
(2.16)
ck := ck (a) :=
ϕ(a)
a
X
w≥1
gcd(w,a)=1
µ2 (w) gcd(w, k)
.
wϕ(w)
Thus, we have
(2.17)
X
d≤x
gcd(d,a)=1
kd
= ck x + O(log(kx)).
ϕ(kd)
!
8
ADAM TYLER FELIX
Note that
1 X kd
1
=
.
ϕ(kd)
k d≤x dϕ(kd)
X
(2.18)
d≤x
gcd(d,a)=1
However, by partial summation, we have
Z x
X
1
1 X
kd
kd
=
+
2
dϕ(kd)
x d≤x ϕ(kd)
1 u
d≤x
gcd(d,a)=1
gcd(d,a)=1
X
d≤u
gcd(d,a)=1
kd
du
ϕ(kd)
Z x
Z x
log(kx)
ck
log(ku)
= ck + O
+
du + O
du
x
u2
1 u
1
= ck log x + O(log k)
(2.19)
since f (w) := gcd(w, k) is a multiplicative function implies that we have
Y
X µ2 (w) gcd(w, k) Y 1
p2 (p − 1)
ck (a) ≤
=
1+
wϕ(w)
p(p − 1)
(p − 1)(p2 − p + 1)
p
w≥1
p|k
Y
Y
1
p−1
≤
1+
1+
p(p − 1) + 1
p−1
p|k
p|k
(2.20)
=
X µ2 (d)
ϕ(d)
d|k
X 1
log k.
ϕ(d)
d≤k
So, we have
X
(2.21)
d≤x
gcd(d,a)=1
1
1
ck
= (ck log x + O(log k)) = log x + O
ϕ(kd)
k
k
log k
k
.
Thus,
X
√x
d≤2
X
π(x; kd, a) = li(x)
√
d≤2 x
k
gcd(d,a)=1
k
1
+O
ϕ(kd)
x
(log x)A
ck
ck log k
ck
x log k
x
=
x−
li(x) +
log 2 + O
+O
2k
2k
2k
k log x
(log x)A
ck
x(log k)(1 + ck )
x
=
x+O
+O
.
2k
k log x
(log x)A
(2.22)
Thus,
(2.23)
X
p≤x
p≡a mod k
d
p−a
k
ck
= x+O
k
x(log k)(1 + ck )
k log x
+O
x
(log x)A
.
GENERALIZING THE TITCHMARSH DIVISOR PROBLEM
9
So we just need to show that the Euler products are true. To see this note that f (w) :=
gcd(w, k) is a multiplicative function. Therefore,
!
1 Y X µ(pn ) gcd(pn , k)
ϕ(a) X µ2 (w) gcd(w, k) Y
=
1−
ck (a) =
a
wϕ(w)
p
pn ϕ(pn )
w≥1
n≥0
p|a
p-a
gcd(w,a)=1
Y
1 Y
gcd(p, k)
=
1−
1+
p
p(p − 1)
p-a
p|a
Y
Y
1 Y
1
1
=
1−
1+
1+
p
p−1
p(p − 1)
p|a
(2.24)
p|k
p-a
p-k
Y
ζ(2)ζ(3) Y
p
p−1
=
1− 2
1+ 2
.
ζ(6)
p −p+1
p −p+1
p|a
p|k
Therefore, Theorem 1.2 holds.
We note that we obtain an improvement in this theorem by using Fiorilli’s work [4, Theorem
3.4] on extending Bombieri, Friedlander, and Iwaniec’s work [1]. This will give us improved
results in the next section.
3. Application to a Generalization of Artin’s Conjecture
Recall for p - a, the order of a modulo p is
(3.1)
fa (p) = min{d ∈ N : ad ≡ 1 mod p},
and for convenience, fa (p) = ∞ for p|a.
We want to consider
X 1
X ia (p)
(3.2)
=
.
f
(p)
p
−
1
a
p≤x
p≤x
We note that if the above summation is x1/4 , then there are infinitely many primes p for
which a is a primitive root (see Murty and Srinivasan [12]). We will show that, on average,
the above summation is log x.
Let us consider the following summation:
1 XX 1
(3.3)
.
y a≤y p≤x fa (p)
We will also see that this summation is related to the Titchmarsh divisor problem (see [2,
§9.3]):
X
x
(3.4)
d(p − a) = cx + c1 li(x) + O
(log x)A
p≤x
10
ADAM TYLER FELIX
for any A > 0, where c and c1 are constants, and the implied constant is dependent only on
a and A. More precisely, this is related to the summation
X
p−1
d
k
p≤x
p≡1 mod k
discussed in the previous section.
l m
y+1
Since fkp (p) = ∞ for any k ∈ Z and y = y+1
p − 1, where dxe is the smallest
p
−
1
≤
p
p
integer greater than x ∈ R, we have


XX 1
X X
X
1
1 
≤
+ ··· +


f
(p)
f
(p)
f
(p)
a
a
a
y+1
y+1
p≤x a≤y
p≤x
1≤a≤p−1
(d p e−1)p+1≤a≤d p ep−1
X y+1 X 1
(3.5)
=
p
f (p)
a≤p−1 a
p≤x
l m
since fa (p) = fa+kp (p) by definition. Therefore, since y+1
= y+1
+ O(1), we have
p
p
X
X y + 1
XX 1
1
≤
+ O(1)
f (p)
p
f (p)
p≤x
a≤p−1 a
p≤x a≤y a
!
X1 X 1
X X 1
(3.6)
= (y + 1)
+O
.
p
f
(p)
f
(p)
a
a
p≤x
a≤p−1
p≤x a≤p−1
Similarly, using [x] the greatest integer smaller than x and truncating the above summations
to a ≤ [(y + 1)/p] instead of extending them to a ≤ d(y + 1)/pe, we have
!
XX 1
X1 X 1
X X 1
.
(3.7)
≥ (y + 1)
+O
f (p)
p a≤p−1 fa (p)
f (p)
p≤x a≤y a
p≤x
p≤x a≤p−1 a
In particular,
XX
p≤x a≤y
(3.8)
X1 X
1
1
= (y + 1)
+O
fa (p)
p a≤p−1 fa (p)
p≤x
1
f (p)
p≤x a≤p−1 a
!
X1 X 1
X X 1
=y
+O
p a≤p−1 fa (p)
f (p)
p≤x
p≤x a≤p−1 a
!
X X
Let us consider the error term first. We note that the number of elements of (Z/pZ)∗ that
have order k is ϕ(k) provided k|p − 1 and 0 otherwise. So, we have
X 1
X #{a ∈ (Z/pZ)∗ : fa (p) = k}
X ϕ(k)
(3.9)
=
=
.
f (p)
k
k
a≤p−1 a
k|p−1
k|p−1
GENERALIZING THE TITCHMARSH DIVISOR PROBLEM
Therefore,
X X ϕ(k) X
1
=
≤
d(p − 1) x
f (p) p≤x
k
p≤x a≤p−1 a
p≤x
X X
(3.10)
k|p−1
by the Titchmarsh divisor problem [2, §9.3].
Let us now consider the main term. We first note that we have
X ϕ(k)
1
=
p a≤p−1 fa (p) k≤x k
X1 X
(3.11)
p≤x
X
p≤x
p≡1 mod k
1
p
since
X ϕ(k)
1
=
.
f
(p)
k
a
a≤p−1
X
(3.12)
k|p−1
By partial summation, we have
X
(3.13)
p≤x
p≡1 mod k
π(x; k, 1)
1
=
+
p
x
Z
k
x
π(u; k, 1)
du.
u2
Therefore,
(3.14)
X1 X
p≤x
X ϕ(k)
1
=
p a≤p−1 fa (p) k≤x k
π(x; k, 1)
+
x
Z
x
k
π(u; k, 1)
du .
u2
We will evaluate each of these sums separately.
The first summation is dealt with as follows:
(3.15)
X ϕ(k) π(x; k, 1)
k≤x
k
x
≤
1X
1X
π(x; k, 1) =
d(p − 1) 1
x k≤x
x p≤x
by the Titchmarsh divisor problem [2, §9.3].
In order to evaluate the second summation notice that
Z x
X ϕ(k) Z x π(u; k, 1)
1 X ϕ(k)
(3.16)
du
=
π(u; k, 1) du.
2
k
u2
k
k
2 u k≤u
k≤x
So, in order to evaluate our desired sum, we need to evaluate
(3.17)
X ϕ(k)
k≤x
k
π(x; k, 1).
11
12
ADAM TYLER FELIX
We have
X ϕ(k)
k≤x
k
π(x; k, 1) =
X X X µ(m) X µ(m)
=
m
m
p≤x
m≤x
k|p−1 m|k
=
X µ(k)
k≤x
k
k≤(log x)A+2
(3.18)
d
p≤x
p≡1 mod k
X
=
X
µ(k)
k
p−1
k
+
d
p−1
k
µ(k)
k
X
p≤x
p≡1 mod k
X
(log x)A+2 <k≤x
X
1
p≤x
k|p−1:
p≡1 mod m m|k
X
X
d
p≤x
p≡1 mod k
p−1
k
.
Let us consider the second summation above. We have
X
(log x)A+2 <k≤x
µ(k)
k
X
p≤x
p≡1 mod k
d
p−1
k
X
(log x)A+2 <k≤x
≤
X
(log x)A+2 <k≤x
(3.19)
x
(log x)A+1
x
.
(log x)A
1
k
1
k
X
d
p≤x
p≡1 mod k
X
d(n)
n≤x/(log x)A+2
X
(log x)A+2 <k≤x
p−1
k
1
k
GENERALIZING THE TITCHMARSH DIVISOR PROBLEM
13
For the first summation, we have
X µ(k) X
p−1
d
k
k
A+2
p≤x
k≤(log x)
p≡1 mod k
ck
x(log k)(1 + ck )
x
x+O
+O
=
k
k log x
(log x)A+1
k≤(log x)A+2


X
X
µ(k)ck
(1 + ck ) log k 
x
=x
+O
2
k
log x
k2
k≤(log x)A+2
k≤(log x)A+2


X
x
1
+O
(log x)A+1
k
k≤(log x)A+2


X
X
µ(k)ck
x
(1 + ck ) log k 
=x
+O
2
k
log x
k2
k≤(log x)A+2
k≤(log x)A+2
x
+O
(log x)A
X
(3.20)
µ(k)
k
by Theorem 1.2. Also,
X µ(k)ck X µ(k)ck X µ(k)ck
=
−
k2
k2
k2
k≥1
k>x
k≤x
∞
∞
X
µ(k) X µ2 (w) gcd(w, k)
∞
X µ(k) X
µ2 (w) gcd(w, k)
−
wϕ(w)
k 2 w=1
k>x


X
∞
Y
Y
1
1


µ(k)
=
1+
2
p(p
−
1)
p
−
p
+
1
p
k=1
p|k


X
Y
1

+O
|µ(k)|
p(p
−
1)
k>x
=
k=1
k2
w=1
wϕ(w)
p|k
!
X |µ(k)|
1
=
+O
p2 − p + 1
kϕ(k)
p
p
k>x
Y (p(p − 1) + 1)(p2 − p + 1 − 1)
1
=
+O
2
p(p − 1)(p − p + 1)
x
p
1
=1+O
.
x
Y
(3.21)
1
1+
p(p − 1)
Y
1−
14
ADAM TYLER FELIX
This proves Lemma 1.1.
As ck log k, we have
X
(3.22)
k≤(log x)A+2
(1 + ck ) log k
1.
k2
So,
X ϕ(k)
(3.23)
k≤x
k
π(x; k, 1) = x 1 + O
(3.24)
=x+O
1
(log x)A+2
+O
x
log x
x
log x
as A > 0 can be chosen to be arbitrarily large. Therefore, Theorem 1.3 holds.
It can now be shown that
x
1 XX 1
= log x + O(log log x) + O
(3.25)
y a≤y p≤x fa (p)
y
assuming
x
log x
(3.26)
= o(y). To see this, we need to evaluate
Z x
1 X ϕ(k)
π(u; k, 1) du.
2
k
2 u k≤u
We have
Z
x
2
Z x
1 X ϕ(k)
1
u
π(u; k, 1) du =
u+O
du
2
u2 k≤u k
log u
2 u
(3.27)
= log x + O(log log x)
by Theorem 1.3. This proves Theorem 1.4 since logx x = o(y) forces xy = o(log x) and so our
summation becomes
1 XX 1
x
(3.28)
= log x + O(log log x) + O
.
y a≤y p≤x fa (p)
y
To see that Theorem 1.5 holds, note that all of the previous error terms and justifications
work for this case as lwell.m To see this, consider the following: since fkp (p) = ∞ for any k ∈ Z
y+1
p
p
(3.29)
−1≤
y+1
p
p − 1, we have

XX
X X
1
1
≤
+ ··· +

ϕ(fa (p))
ϕ(fa (p))
p≤x a≤y
p≤x
1≤a≤p−1
and y =
=
X y + 1 X
p≤x
p
1
ϕ(fa (p))
a≤p−1

X
−1)p+1≤a≤d y+1
p−1
(d y+1
p e
p e
1


ϕ(fa (p))
GENERALIZING THE TITCHMARSH DIVISOR PROBLEM
l
m
since fa (p) = fa+kp (p) by definition. Therefore, since
= y+1
+ O(1), we have
p
X y+1
X
XX
1
1
≤
+ O(1)
ϕ(fa (p))
p
ϕ(fa (p))
p≤x
a≤p−1
p≤x a≤y
X1 X
1
= (y + 1)
p a≤p−1 ϕ(fa (p))
p≤x
!
X X
1
(3.30)
.
+O
ϕ(f
(p))
a
p≤x a≤p−1
Similarly, we have
XX
p≤x a≤y
y+1
p
15
X1 X
1
1
≥ (y + 1)
ϕ(fa (p))
p a≤p−1 ϕ(fa (p))
p≤x
+O
(3.31)
1
ϕ(fa (p))
p≤x a≤p−1
!
X X
.
In particular,
XX
p≤x a≤y
(3.32)
X1 X
X X
1
1
1
= (y + 1)
+O
ϕ(fa (p))
p a≤p−1 ϕ(fa (p))
ϕ(fa (p))
p≤x
p≤x a≤p−1
!
X1 X
X X
1
1
+O
=y
p a≤p−1 ϕ(fa (p))
ϕ(fa (p))
p≤x
p≤x a≤p−1
!
Since
(3.33)
X #{1 ≤ a ≤ p − 1 : fa (p) = k}
X
1
=
=
1 = d(p − 1),
ϕ(fa (p))
ϕ(k)
a≤p−1
X
k|p−1
k|p−1
we have
(3.34)
X
1
=
d(p − 1) x
ϕ(f
(p))
a
p≤x a≤p−1
p≤x
X X
by the Titchmarsh divisor problem.
Similarly,
X1 X
X π(x; k, 1) Z x π(u; k, 1) 1
(3.35)
=
+
du .
2
p
ϕ(f
(p))
x
u
a
k
p≤x
a≤p−1
k≤x
The first summation is bounded by 1 using the Titchmarsh Divisor problem as before. The
integral becomes
Z x
X Z x π(u; k, 1)
1 X
(3.36)
du
=
π(u; k, 1) du.
2
u2
2 u k≤u
k≤x k
16
ADAM TYLER FELIX
However, this inner summation becomes
(3.37)
u
ζ(2)ζ(3)
u + c1
+O
π(u; k, 1) =
d(p − 1) =
ζ(6)
log
u
p≤u
k≤u
X
X
u
(log u)2
.
Hence,
(3.38)
XZ
k≤x
k
x
π(u; k, 1)
ζ(2)ζ(3)
log x + c1 log log x + O(1).
du
=
u2
ζ(6)
Therefore, Theorem 1.5 holds.
Remarks
We note that in both of Theorems 1.4 and 1.5, there is a log log x term. We also have
(3.39)
X
p≤x
X ia (p)
1
=
.
fa (p) p≤x p − 1
We could apply partial summation the right-hand side but an estimate on the summation
X
ia (p)
p≤x
would be needed. Currently, if we assume GRH for the Dedekind zeta functions of Q(ζn , a1/n )
as n ranges over N, we get lower bounds of the form
X
(3.40)
ia (p) x
p≤x
(see [3, Chapter 7, §1]). Unconditionally, we have
(3.41)
X
p≤x
ia (p) x log log x
.
log x
To see this let πd (x) := #{p ≤ x : d|ia (p)}. Then, by [8, Page 213], d|ia (p) if and only if p splits
completely in Q(ζd , a1/d ) where ζd is a primitive dth root of unity. So, by the unconditional
effective Chebotarev density theorem ([9, Theorem 1.4] or [13, Page 376]), we have for any
GENERALIZING THE TITCHMARSH DIVISOR PROBLEM
17
A>1
X
ia (p) =
p≤x
X
ϕ(d)πd (x) ≥
d≤x
X
ϕ(d)πd (x)
d≤(log x)1/7
x
li(x)
+O
=
ϕ(d)
[Kd : Q]
(log x)A
1/7
d≤(log x)
X 1
x
+O
(by [16, Proposition 4.1])
li(x)
d
(log x)A
1/7
X
d≤(log x)
x log log x
.
log x
The belief [5, Conjecture 1(a)] is that we have
X
(3.43)
ia (p) ∼ ca x
(3.42)
p≤x
where ca is a positive constant dependent on a. Theorem 1.4 indicates that we may have
X
x
(3.44)
ia (p) = ca x + O
.
log x
p≤x
We note that in Theorem 1.4 and 1.5, there is a log log x term. In fact, in Theorem 1.5, this
term contributes to the summation. This suggests that we may have
X
x
(3.45)
ia (p) = ca x + Ω
.
log x
p≤x
As mentioned in the previous section, Fiorilli [4, Theorem 3.4] allow us to improve Theorem
1.4 by replacing O(log log y) by c log log y + O(1) where c is a constant. We relegate further
analysis and computation of this summation to future research.
Acknowledgements
Some portions of this work were part of the doctoral thesis of the author [3]. The author
would like Dr. M. Ram Murty, his supervisor, for comments on previous versions of this paper
and guidance throughout the author’s graduate school years at Queen’s University. The author
wishes to thank Dr. Amir Akbary and the reviewer for comments on previous versions of this
paper. The author also wishes to thank Daniel Fiorilli for his comments and his preprint [4].
References
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ADAM TYLER FELIX
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141–150.
Department of Mathematics, Queen’s University, Kingston, Ontario, K7L 3N6, Canada
E-mail address: [email protected]